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C Integ Progr 58257 pair #381736291
details
property
value
status
complete
benchmark
PastaB2.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n025.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
1.9097328186 seconds
cpu usage
4.883857419
max memory
3.81845504E8
stage attributes
key
value
output-size
2299
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_c /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.c # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 49 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 0 ms] (6) IntTRS (7) RankingReductionPairProof [EQUIVALENT, 23 ms] (8) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f1(x, y) -> f2(x_1, y) :|: TRUE f2(x1, x2) -> f3(x1, x3) :|: TRUE f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 - 1 f5(x14, x15) -> f6(x14, x16) :|: TRUE && x16 = x15 + 1 f3(x8, x9) -> f4(x8, x9) :|: x8 > x9 f6(x10, x11) -> f3(x10, x11) :|: TRUE f3(x12, x13) -> f7(x12, x13) :|: x12 <= x13 Start term: f1(x, y) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f3(x8, x9) -> f4(x8, x9) :|: x8 > x9 f6(x10, x11) -> f3(x10, x11) :|: TRUE f5(x14, x15) -> f6(x14, x16) :|: TRUE && x16 = x15 + 1 f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 - 1 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f5(x14:0, x15:0) -> f5(x14:0 - 1, x15:0 + 1) :|: x15:0 + 1 < x14:0 ---------------------------------------- (7) RankingReductionPairProof (EQUIVALENT) Interpretation: [ f5 ] = -1/2*f5_2 + 1/2*f5_1 The following rules are decreasing: f5(x14:0, x15:0) -> f5(x14:0 - 1, x15:0 + 1) :|: x15:0 + 1 < x14:0 The following rules are bounded: f5(x14:0, x15:0) -> f5(x14:0 - 1, x15:0 + 1) :|: x15:0 + 1 < x14:0 ---------------------------------------- (8) YES
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